Question

Suppose you were to wrap the Moon in aluminum foil and place a charge \(Q\) on it. What is the capacitance of the Moon in this case? [Hint: It is not necessary to have two oppositely charged conductors to have a capacitor. Use the definition of potential for a spherical conductor and the definition of capacitance to get your answer. \(]\)

Short answer

Answer: The capacitance of the Moon wrapped in aluminum foil and charged with Q is approximately 1.94 x 10^-4 F.

Step-by-step solution

1. Define the capacitance of a capacitor

The capacitance (\(C\)) of a capacitor is defined by the relation: \(C = \frac{Q}{V}\), where \(Q\) is the charge on the plates and \(V\) is the potential difference between the plates.

2. Calculate the potential of the Moon as a spherical conductor

The potential of a spherical conductor with a charge \(Q\) and radius \(R\) can be given by the formula: \(V = \frac{kQ}{R}\), where \(k\) is the electrostatic constant, which is approximately \(8.99 \times 10^9 \, Nm^2/C^2\). The radius of the Moon, \(R\), is about \(1.74 \times 10^6 \, m\) and we are given the charge \(Q\).

3. Apply the capacitance formula

From step 1, we know that \(C = \frac{Q}{V}\). Now, substitute the formula for potential from step 2: \(C = \frac{Q}{\frac{kQ}{R}}\). When we simplify this expression, we get: \(C = \frac{Q}{kQ} \times R = \frac{1}{k} \times R\).

4. Calculate the capacitance

Now that we have the formula for capacitance in terms of the radius of the Moon and the electrostatic constant, we can calculate the capacitance: \(C = \frac{1}{k} \times R = \frac{1}{8.99 \times 10^9 \, Nm^2/C^2} \times 1.74 \times 10^6 \, m\). After the calculation, we have: \(C \approx 1.94 \times 10^{-4} \, F\). Finally, the capacitance of the Moon wrapped in aluminum foil and charged with \(Q\) is approximately \(1.94 \times 10^{-4} \, F\).